Alg7.4

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Prove that if x^{2}=1\forall x\in G\,, then G\, is abelian.

(1)\forall x\in G,x\cdot x=1\implies x=x^{{-1}}\,

If a,b\in G,(aa)(bb)=1\,

a^{{-1}}aabbb^{{-1}}=a^{{-1}}b^{{-1}}\,

ab=a^{{-1}}b^{{-1}}=(ba)^{{-1}}\,

But from (1),ba=(ba)^{{-1}}\,

So now ab=ba\,



Or, even simpler:


ab=1ab1=(bb)ab(aa)=b(ba)(ba)a=b1a=ba\,

Since this proof doesn't use inverses, the property holds for all monoids, not only groups.

But every monoid, for which this property holds, is in fact a group.

Abstract Algebra

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